This chapter discusses vibration dynamics and methods for deriving equations of motion. The Newton-Euler and Lagrange methods are commonly used to derive equations of motion for vibrating systems. The Newton-Euler method is well-suited for discrete, lumped parameter models with a low degree of freedom. It involves drawing free body diagrams and applying Newton's second law to each mass to obtain the equations of motion. Having symmetric coefficient matrices is the main advantage of using the Lagrange method for mechanical vibrations.

1. Chapter 2
Vibration Dynamics
In this chapter, we review the dynamics of vibrations and the methods of deriv-
ing the equations of motion of vibrating systems. The Newton–Euler and Lagrange
methods are the most common methods of deriving the equations of motion. Having
symmetric coefficient matrices is the main advantage of using the Lagrange method
in mechanical vibrations.
Fig. 2.1 A one DOF
vibrating systems and its FBD
2.1 Newton–Euler Method
When a vibrating system is modeled as a combination of masses mi , dampers ci ,
and springs ki , it is called a discrete or lumped model of the system.
To find the equations of motion of a low degree-of-freedom (DOF) discrete
model of a vibrating system, the Newton–Euler method works very well. We move
˙
all the masses mi out of their equilibria at positions xi with velocities xi . Then a free
body diagram (FBD) of the lumped masses indicates the total force Fi on mass mi .
R.N. Jazar, Advanced Vibrations, 51
DOI 10.1007/978-1-4614-4160-1_2, © Springer Science+Business Media New York 2013

2. 52 2 Vibration Dynamics
Employing the momentum pi = mi vi of the mass mi , the Newton equation provides
us with the equation of motion of the system:
d d
Fi = pi = (mi vi ) (2.1)
dt dt
When the motion of a massive body with mass moment Ii is rotational, then its equa-
tion of motion will be found by Euler equation, in which we employ the moment of
momentum Li = Ii ω of the mass mi :
d d
Mi = Li = (Ii ω) (2.2)
dt dt
For example, Fig. 2.1 illustrates a one degree-of-freedom (DOF) vibrating sys-
tem. Figure 2.1(b) depicts the system when m is out of the equilibrium position at x
˙
and moving with velocity x, both in positive direction. The FBD of the system is as
shown in Fig. 2.1(c). The Newton equation generates the equations of motion:
ma = −cv − kx + f (x, v, t) (2.3)
The equilibrium position of a vibrating system is where the potential energy of
the system, V , is extremum:
∂V
=0 (2.4)
∂x
We usually set V = 0 at the equilibrium position. Linear systems with constant
stiffness have only one equilibrium or infinity equilibria, while nonlinear systems
may have multiple equilibria. An equilibrium is stable if
∂ 2V
>0 (2.5)
∂x 2
and is unstable if
∂ 2V
<0 (2.6)
∂x 2
The geometric arrangement and the number of employed mechanical elements
can be used to classify discrete vibrating systems. The number of masses times the
DOF of each mass makes the total DOF of the vibrating system n. Each indepen-
dent DOF of a mass is indicated by an independent variable, called the generalized
coordinate. The final set of equations would be n second-order differential equa-
tions to be solved for n generalized coordinates. When each moving mass has one
DOF, then the system’s DOF is equal to the number of masses. The DOF may
also be defined as the minimum number of independent coordinates that defines the
configuration of a system.
The equation of motion of an n DOF linear mechanical vibrating system of can
always be arranged as a set of second-order differential equations
[m]¨ + [c]˙ + [k]x = F
x x (2.7)

3. 2.1 Newton–Euler Method 53
Fig. 2.2 Two, three, and one
DOF models for vertical
vibrations of vehicles
Fig. 2.3 A 1/8 car model
and its free body diagram
in which, x is a column array of describing coordinates of the system, and f is a
column array of the associated applied forces. The square matrices [m], [c], [k] are
the mass, damping, and stiffness matrices.
Example 30 (The one, two, and three DOF model of vehicles) The one, two, and
three DOF model for analysis of vertical vibrations of a vehicle are shown in
Fig. 2.2(a)–(c). The system in Fig. 2.2(a) is called the quarter car model, in which
ms represents a quarter mass of the body, and mu represents a wheel. The param-
eters ku and cu are models for tire stiffness and damping. Similarly, ks and cu are
models for the main suspension of the vehicle. Figure 2.2(c) is called the 1/8 car
model, which does not show the wheel of the car, and Fig. 2.2(b) is a quarter car
with a driver md . The driver’s seat is modeled by kd and cd .
Example 31 (1/8 car model) Figure 2.3(a) shows the simplest model for vertical
vibrations of a vehicle. This model is sometimes called 1/8 car model. The mass ms
represents one quarter of the car’s body, which is mounted on a suspension made of
a spring ks and a damper cs . When ms is at a position such as shown in Fig. 2.3(b),
its free body diagram is as in Fig. 2.3(c).

4. 54 2 Vibration Dynamics
Fig. 2.4 Equivalent
mass–spring vibrator for a
pendulum
Applying Newton’s method, the equation of motion would be
ms x = −ks (xs − y) − cs (xs − y)
¨ ˙ ˙ (2.8)
which can be simplified to
ms x + cs xs + ks xs = ks y + cs y
¨ ˙ ˙ (2.9)
The coordinate y indicates the input from the road and x indicates the absolute
displacement of the body. Absolute displacement refers to displacement with respect
to the motionless background.
Example 32 (Equivalent mass and spring) Figure 2.4(a) illustrates a pendulum made
by a point mass m attached to a massless bar with length l. The coordinate θ shows
the angular position of the bar. The equation of motion for the pendulum can be
found by using the Euler equation and employing the FBD shown in Fig. 2.4(b):
¨
IA θ = MA (2.10)
¨
ml 2 θ = −mgl sin θ (2.11)
Simplifying the equation of motion and assuming a very small swing angle yields
¨
l θ + gθ = 0 (2.12)
This equation is equivalent to an equation of motion of a mass–spring system made
by a mass me ≡ l, and a spring with stiffness ke ≡ g. The displacement of the mass
would be xe ≡ θ . Figure 2.4(c) depicts such an equivalent mass–spring system.
Example 33 (Gravitational force in rectilinear vibrations) When the direction of
the gravitational force on a mass m is not varied with respect to the direction of

5. 2.1 Newton–Euler Method 55
Fig. 2.5 A mass–spring–damper system indicating that the gravitational force in rectilinear vibra-
tions provides us with a static deflection
motion of m, the effect of the weight force can be ignored in deriving the equation
of motion. In such a case the equilibrium position of the system is at a point where
the gravity is in balance with a deflection in the elastic member. This force–balance
equation will not be altered during vibration. Consequently we may ignore both
forces; the gravitational force and the static elastic force. It may also be interpreted
as an energy balance situation where the work of gravitational force is always equal
to the extra stored energy in the elastic member.
Consider a spring k and damper c as is shown in Fig. 2.5(a). A mass m is put on
the force free spring and damper. The weight of m compresses the spring a static
length xs to bring the system at equilibrium in Fig. 2.5(b). When m is at equilibrium,
it is under the balance of two forces, mg and −kxs :
mg − kxs = 0 (2.13)
While the mass is in motion, its FBD is as shown in Fig. 2.5(c) and its equation of
motion is
mx = −kx − cx + mg − kxs
¨ ˙
= −kx − cx
˙ (2.14)
It shows that if we examine the motion of the system from equilibrium, we can
ignore both the gravitational force and the initial compression of the elastic member
of the system.
Example 34 (Force proportionality) The equation of motion for a vibrating system
is a balance between four different forces: a force proportional to displacement,
−kx, a force proportional to velocity, −cv, a force proportional to acceleration, ma,
and an applied external force f (x, v, t), which can be a function of displacement,
velocity, and time. Based on Newton method, the force proportional to acceleration,
ma, is always equal to the sum of all the other forces:
ma = −cv − kx + f (x, v, t) (2.15)

6. 56 2 Vibration Dynamics
Fig. 2.6 A 1/4 car model
and its free body diagram
Example 35 (A two DOF base excited system) Figure 2.6(a)–(c) illustrate the equi-
librium, motion, and FBD of a two DOF system. The FBD is plotted based on the
assumption
xs > x u > y (2.16)
Applying Newton’s method provides us with two equations of motion:
ms xs = −ks (xs − xu ) − cs (xs − xu )
¨ ˙ ˙ (2.17)
mu xu = ks (xs − xu ) + cs (xs − xu )
¨ ˙ ˙
− ku (xu − y) − cu (xu − y)
˙ ˙ (2.18)
The assumption (2.16) is not necessary. We can find the same Eqs. (2.17)
and (2.18) using any other assumption, such as xs < xu > y, xs > xu < y, or
xs < xu < y. However, having an assumption helps to make a consistent free body
diagram.
We usually arrange the equations of motion for a linear system in a matrix form
to take advantage of matrix calculus:
[M]˙ + [C]˙ + [K]x = F
x x (2.19)
Rearrangement of Eqs. (2.17) and (2.18) yields
ms 0 ¨
xs cs −cs ˙
xs
+
0 mu ¨
xu −cs cs + cu ˙
xu
ks −ks xs 0
+ = (2.20)
−ks ks + ku xu ku y + c u y
˙
Example 36 (Inverted pendulum and negative stiffness) Figure 2.7(a) illustrates
an inverted pendulum with a tip mass m and a length l. The pendulum is supported

7. 2.1 Newton–Euler Method 57
Fig. 2.7 An inverted
pendulum with a tip mass m
and two supportive springs
by two identical springs attached to point B at a distance a < l from the pivot A.
A free body diagram of the pendulum is shown in Fig. 2.7(b). The equation of
motion may be found by taking a moment about A:
¨
MA = I A θ (2.21)
¨
mg(l sin θ ) − 2kaθ (a cos θ ) = ml 2 θ (2.22)
To derive Eq. (2.22) we assumed that the springs are long enough to remain almost
straight when the pendulum oscillates. Rearrangement and assuming a very small θ
show that the nonlinear equation of motion (2.22) can be approximated by
¨
ml 2 θ + mgl − 2ka 2 θ = 0 (2.23)
which is equivalent to a linear oscillator:
¨
me θ + ke θ = 0 (2.24)
with an equivalent mass me and equivalent stiffness ke :
me = ml 2 ke = mgl − 2ka 2 (2.25)
The potential energy of the inverted pendulum is
V = −mgl(1 − cos θ ) + ka 2 θ 2 (2.26)
which has a zero value at θ = 0. The potential energy V is approximately equal to
the following equation if θ is very small:
1
V ≈ − mglθ 2 + ka 2 θ 2 (2.27)
2

8. 58 2 Vibration Dynamics
because
1
cos θ ≈ 1 − θ 2 + O θ 4 (2.28)
2
To find the equilibrium positions of the system, we may solve the equation ∂V /∂θ =
0 for any possible θ :
∂V
= −2mglθ + 2ka 2 θ = 0 (2.29)
∂θ
The solution of the equation is
θ =0 (2.30)
which shows that the upright vertical position is the only equilibrium of the inverted
pendulum as long as θ is very small. However, if
mgl = ka 2 (2.31)
then any θ around θ = 0 would be an equilibrium position and, hence, the inverted
pendulum would have an infinity of equilibria.
The second derivative of the potential energy
∂ 2V
= −2mgl + 2ka 2 (2.32)
∂x 2
indicates that the equilibrium position θ = 0 is stable if
ka 2 > mgl (2.33)
A stable equilibrium pulls the system back if it deviates from the equilibrium, while
an unstable equilibrium repels the system. Vibration happens when the equilibrium
is stable.
This example also indicates the fact that having a negative stiffness is possible
by geometric arrangement of mechanical components of a vibrating system.
Example 37 (Force function in equation of motion) Qualitatively, force is what-
ever changes the motion, and quantitatively, force is whatever is equal to mass times
acceleration. Mathematically, the equation of motion provides us with a vectorial
second-order differential equation
m¨ = F(˙ , r, t)
r r (2.34)
We assume that the force function may generally be a function of time t, position r,
˙
and velocity r. In other words, the Newton equation of motion is correct as long as
we can show that the force is only a function of r, r, t.˙
If there is a force that depends on the acceleration, jerk, or other variables that
˙
cannot be reduced to r, r, t, the system is not Newtonian and we do not know the
equation of motion, because
...
˙ ¨
F(r, r, r, r , . . . , t) = m¨
r (2.35)

9. 2.2 Energy 59
˙
In Newtonian mechanics, we assume that force can only be a function of r, r, t and
nothing else. In real world, however, force may be a function of everything; however,
˙
we always ignore any other variables than r, r, t.
Because Eq. (2.34) is a linear equation for force F, it accepts the superposition
principle. When a mass m is affected by several forces F1 , F2 , F3 , . . . , we may
calculate their summation vectorially
F = F1 + F2 + F3 + · · · (2.36)
and apply the resultant force on m. So, if a force F1 provides us with acceleration
¨ ¨
r1 , and F2 provides us with r2 ,
m¨ 1 = F1
r m¨ 2 = F2
r (2.37)
¨
then the resultant force F3 = F1 + F2 provides us with the acceleration r3 such that
¨ ¨ ¨
r3 = r1 + r 2 (2.38)
To see that the Newton equation of motion is not correct when the force is not
˙
only a function of r, r, t, let us assume that a particle with mass m is under two
¨ ¨
acceleration dependent forces F1 (x) and F2 (x) on x-axis:
¨
mx1 = F1 (x1 )
¨ mx2 = F2 (x2 )
¨ ¨ (2.39)
¨
The acceleration of m under the action of both forces would be x3
mx3 = F1 (x3 ) + F2 (x3 )
¨ ¨ ¨ (2.40)
however, though we must have
¨
x3 = x1 + x2
¨ ¨ (2.41)
we do have
¨
m(x1 + x2 ) = F1 (x1 + x2 ) + F2 (x1 + x2 )
¨ ¨ ¨ ¨ ¨
= F1 (x1 ) + F2 (x2 )
¨ ¨ (2.42)
2.2 Energy
In Newtonian mechanics, the acting forces on a system of bodies can be divided
into internal and external forces. Internal forces are acting between bodies of the
system, and external forces are acting from outside of the system. External forces
and moments are called the load. The acting forces and moments on a body are
called a force system. The resultant or total force F is the sum of all the external
forces acting on the body, and the resultant or total moment M is the sum of all

10. 60 2 Vibration Dynamics
the moments of the external forces about a point, such as the origin of a coordinate
frame:
F= Fi M= Mi (2.43)
i i
The moment M of a force F, acting at a point P with position vector rP , about a
point Q at rQ is
MQ = (rP − rQ ) × F (2.44)
and, therefore, the moment of F about the origin is
M = rP × F (2.45)
The moment of the force about a directional line l passing through the origin is
Ml = u · (rP × F)
ˆ (2.46)
ˆ
where u is a unit vector on l. The moment of a force may also be called torque or
moment.
The effect of a force system is equivalent to the effect of the resultant force and
resultant moment of the force system. Any two force systems are equivalent if their
resultant forces and resultant moments are equal. If the resultant force of a force
system is zero, the resultant moment of the force system is independent of the origin
of the coordinate frame. Such a resultant moment is called a couple.
When a force system is reduced to a resultant FP and MP with respect to a
reference point P , we may change the reference point to another point Q and find
the new resultants as
FQ = FP (2.47)
MQ = MP + (rP − rQ ) × FP = MP + Q rP × FP (2.48)
The momentum of a moving rigid body is a vector quantity equal to the total mass
of the body times the translational velocity of the mass center of the body:
p = mv (2.49)
The momentum p is also called the translational momentum or linear momentum.
Consider a rigid body with momentum p. The moment of momentum, L, about a
directional line l passing through the origin is
Ll = u · (rC × p)
ˆ (2.50)
ˆ
where u is a unit vector indicating the direction of the line, and rC is the position
vector of the mass center C. The moment of momentum about the origin is
L = rC × p (2.51)
The moment of momentum L is also called angular momentum.

11. 2.2 Energy 61
Kinetic energy K of a moving body point P with mass m at a position G rP , and
having a velocity G vP , in the global coordinate frame G is
1 1
K = mG vP · G vP = mG v2
P (2.52)
2 2
where G indicates the global coordinate frame in which the velocity vector vP is
expressed. The work done by the applied force G F on m in moving from point 1 to
point 2 on a path, indicated by a vector G r, is
2
1 W2 = G
F · dG r (2.53)
1
However,
2 2 Gd 1 2 d 2
G
F · dG r = m G
v · G v dt = m v dt
1 1 dt 2 1 dt
1
= m v2 − v1 = K2 − K1
2 2
(2.54)
2
which shows that 1 W2 is equal to the difference of the kinetic energy of terminal
and initial points:
1 W2 = K2 − K1 (2.55)
Equation (2.55) is called the principle of work and energy. If there is a scalar po-
tential field function V = V (x, y, z) such that
dV ∂V ∂V ∂V ˆ
F = −∇V = − =− ı+
ˆ j+
ˆ k (2.56)
dr ∂x ∂y ∂z
then the principle of work and energy (2.55) simplifies to the principle of conserva-
tion of energy,
K1 + V1 = K2 + V2 (2.57)
The value of the potential field function V = V (x, y, z) is the potential energy of
the system.
Proof Consider the spatial integral of Newton equation of motion
2 2
F · dr = m a · dr (2.58)
1 1
We can simplify the right-hand side of the integral (2.58) by the change of variable
r2 r2 t2 dv
F · dr = m a · dr = m · v dt
r1 r1 t1 dt
v2 1
=m v · dv = m v2 − v2
2 1 (2.59)
v1 2

12. 62 2 Vibration Dynamics
The kinetic energy of a point mass m that is at a position defined by G r and having
a velocity G v is defined by (2.52). Whenever the global coordinate frame G is the
only involved frame, we may drop the superscript G for simplicity. The work done
by the applied force G F on m in going from point r1 to r2 is defined by (2.53).
Hence the spatial integral of equation of motion (2.58) reduces to the principle of
work and energy (2.55):
1 W2 = K2 − K1 (2.60)
which says that the work 1 W2 done by the applied force G F on m during the dis-
placement r2 − r1 is equal to the difference of the kinetic energy of m.
If the force F is the gradient of a potential function V ,
F = −∇V (2.61)
then F · dr in Eq. (2.58) is an exact differential and, hence,
2 2
F · dr = dV = −(V2 − V1 ) (2.62)
1 1
E = K1 + V1 = K2 + V2 (2.63)
In this case the work done by the force is independent of the path of motion between
r1 and r2 and depends only upon the value of the potential V at start and end points
of the path. The function V is called the potential energy; Eq. (2.63) is called the
principle of conservation of energy, and the force F = −∇V is called a potential, or a
conservative force. The kinetic plus potential energy of the dynamic system is called
the mechanical energy of the system and is denoted by E = K + V . The mechanical
energy E is a constant of motion if all the applied forces are conservative.
A force F is conservative only if it is the gradient of a stationary scalar function.
The components of a conservative force will only be functions of space coordinates:
ı ˆ ˆ
F = Fx (x, y, z)ˆ + Fy (x, y, z)j + Fz (x, y, z)k (2.64)
Example 38 (Energy and equation of motion) Whenever there is no loss of energy
in a mechanical vibrating system, the sum of kinetic and potential energies is a
constant of motion:
E = K + V = const (2.65)
A system with constant energy is called a conservative system. The time derivative
of a constant of motion must be zero at all time.
The mass–spring system of Fig. 2.10 is a conservative system with the total me-
chanical energy of
1 1
˙
E = mx 2 + kx 2 (2.66)
2 2

13. 2.2 Energy 63
Fig. 2.8 A multi DOF
conservative vibrating system
Having a zero rate of energy,
˙
E = mx x + kx x = x(mx + kx) = 0
˙¨ ˙ ˙ ¨ (2.67)
˙
and knowing that x cannot be zero at all times provides us with the equation of
motion:
mx + kx = 0
¨ (2.68)
Example 39 (Energy and multi DOF systems) We may use energy method and
determine the equations of motion of multi DOF conservative systems. Consider
the system in Fig. 2.8 whose mechanical energy is
1 1 1
E = K + V = m1 x1 + m2 x2 + m3 x3
˙2 ˙2 ˙2
2 2 2
1 1 1 1
+ k1 x1 + k2 (x1 − x2 )2 + k3 (x2 − x3 )2 + k4 x3 (2.69)
2 2
2 2 2 2
To find the first equation of motion associated to x1 , we assume x2 and x3 are con-
stant and take the time directive:
˙
E = m1 x1 x1 + k1 x1 x1 + k2 (x1 − x2 )x1 = 0
˙ ¨ ˙ ˙ (2.70)
˙
Because x1 cannot be zero at all times, the first equation of motion is
m1 x1 + k1 x1 + k2 (x1 − x2 ) = 0
¨ (2.71)
To find the second equation of motion associated to x2 , we assume that x1 and x3
are constant and we take a time directive of E
˙
E = m2 x2 x2 − k2 (x1 − x2 )x2 + k3 (x2 − x3 )x2 = 0
˙ ¨ ˙ ˙ (2.72)
which provides us with
m2 x2 − k2 (x1 − x2 ) + k3 (x2 − x3 ) = 0
¨ (2.73)
To find the second equation of motion associated to x2 , we assume that x1 and x3
are constant and we take the time directive of E
˙
E = m3 x3 x3 − k3 (x2 − x3 )x3 + k4 x3 x3 = 0
˙ ¨ ˙ ˙ (2.74)

14. 64 2 Vibration Dynamics
Fig. 2.9 A two DOF
conservative nonlinear
vibrating system
which provides us with
m3 x3 − k3 (x2 − x3 ) + k4 x3 = 0
¨ (2.75)
We may set up the equations in a matrix form:
⎡ ⎤⎡ ⎤
m1 0 0 ¨
x1
⎣ 0 m2 0 ⎦ ⎣x2 ⎦ ¨
0 0 m3 ¨
x3
⎡ ⎤⎡ ⎤
k1 + k2 −k2 0 x1
+ ⎣ −k2 k2 + k3 −k3 ⎦ ⎣x2 ⎦ = 0 (2.76)
0 −k3 k3 + k4 x3
Example 40 (Energy and nonlinear multi DOF systems) The energy method can
be applied on every conservative system regardless of linearity of the system. Fig-
ure 2.9 illustrates a two DOF nonlinear system whose kinetic and potential energies
are
1 1
˙ ˙ ˙˙
K = m1 x 2 + m2 x 2 + k 2 θ 2 + 2l x θ sin θ
˙
2
(2.77)
2 2
1
V = kx 2 − m2 g(x − l cos θ ) (2.78)
2
We assumed that the motionless hanging down position is the equilibrium of inter-
est, and that the gravitational energy is zero at the level of m1 at the equilibrium.
Example 41 (Maximum energy and frequency of vibrations) The mechanical vibra-
tions is a continuous exchange of energy between kinetic and potential. If there is
no waste of energy, their maximum values must be equal.
Consider the simple mass–spring system of Fig. 2.10. The harmonic motion,
kinetic energy, and potential energy of the system are
x = X sin ωt (2.79)
1 1
˙
K = mx 2 = mX 2 ω2 cos2 ωt (2.80)
2 2

15. 2.2 Energy 65
Fig. 2.10 A mass–spring
system
Fig. 2.11 A wheel turning,
without slip, over a
cylindrical hill
1 1
V = kx 2 = kX 2 sin2 ωt (2.81)
2 2
Equating the maximum K and V
1 1
mX 2 ω2 = kX 2 (2.82)
2 2
provides us with the frequency of vibrations:
k
ω2 = (2.83)
m
Example 42 (Falling wheel) Figure 2.11 illustrates a wheel turning, without slip,
over a cylindrical hill. We may use the conservation of mechanical energy to find
the angle at which the wheel leaves the hill.
Initially, the wheel is at point A. We assume the initial kinetic and potential, and
hence, the mechanical energies E = K + V are zero. When the wheel is turning
over the hill, its angular velocity, ω, is
v
ω= (2.84)
r
where v is the speed at the center of the wheel. At any other point B, the wheel
achieves some kinetic energy and loses some potential energy. At a certain angle,
where the normal component of the weight cannot provide more centripetal force,
mv 2
mg cos θ = (2.85)
R+r

16. 66 2 Vibration Dynamics
Fig. 2.12 A turning wheel
moving up a step
the wheel separates from the surface. Employing the conservation of energy, we
have
KA + VA = KB + VB (2.86)
The kinetic and potential energy at the separation point B are
1 1
KB = mv 2 + IC ω2 (2.87)
2 2
VB = −mg(R + r)(1 − cos θ ) (2.88)
where IC is the mass moment of inertia for the wheel about its center. Therefore,
1 2 1
mv + IC ω2 = mg(R + r)(1 − cos θ ) (2.89)
2 2
and substituting (2.84) and (2.85) yields
IC
1+ (R + r)g cos θ = 2g(R + r)(1 − cos θ ) (2.90)
mr 2
and, therefore, the separation angle is
2mr 2
θ = cos−1 (2.91)
IC + 3mr 2
Let us examine the equation for a disc wheel with
1
IC = mr 2 (2.92)
2
and find the separation angle:
4
θ = cos−1 ≈ 0.96 rad ≈ 55.15 deg (2.93)
7
Example 43 (Turning wheel over a step) Figure 2.12 illustrates a wheel of radius
R turning with speed v to go over a step with height H < R. We may use the

17. 2.2 Energy 67
principle of energy conservation and find the speed of the wheel after getting across
the step. Employing the conservation of energy, we have
KA + VA = KB + VB (2.94)
1 2 1 1 2 1
mv + IC ω1 + 0 =
2
mv + IC ω2 + mgH
2
(2.95)
2 1 2 2 2 2
IC 2 IC 2
m + 2 v1 = m + 2 v2 + 2mgH (2.96)
R R
and, therefore,
2gH
v2 = v1 −
2
IC
(2.97)
1+ mR 2
The condition for having a real v2 is
2gH
v1 > IC
(2.98)
1+ mR 2
The second speed (2.97) and the condition (2.98) for a solid disc with IC = mR 2 /2
are
4
v2 = v1 − Hg
2 (2.99)
3
4
v1 > Hg (2.100)
3
Example 44 (Newton equation) The application of a force system is emphasized by
Newton’s second law of motion, which states that the global rate of change of linear
momentum is proportional to the global applied force:
Gd Gd
G
F= G
p= mG v (2.101)
dt dt
The second law of motion can be expanded to include rotational motions. Hence, the
second law of motion also states that the global rate of change of angular momentum
is proportional to the global applied moment:
Gd
G
M= G
L (2.102)
dt

18. 68 2 Vibration Dynamics
Proof Differentiating the angular momentum (2.51) shows that
Gd Gd G dr G dp
C
G
L= (rC × p) = × p + rC ×
dt dt dt dt
G dp
= G rC × = G rC × G F = G M (2.103)
dt
Example 45 (Integral and constant of motion) Any equation of the form
˙
f (q, q, t) = c (2.104)
˙
c = f (q0 , q0 , t0 ) (2.105)
q = q1 q2 · · · qn (2.106)
with total differential
n
df ∂f ∂f ∂f
= qi +
˙ qi +
¨ =0 (2.107)
dt ∂qi ˙
∂ qi ∂t
i=1
˙
that the generalized positions q and velocities q of a dynamic system must satisfy
at all times t is called an integral of motion. The parameter c, of which the value
depends on the initial conditions, is called a constant of motion. The maximum
number of independent integrals of motion for a dynamic system with n degrees
of freedom is 2n. A constant of motion is a quantity of which the value remains
constant during the motion.
Any integral of motion is a result of a conservation principle or a combination
of them. There are only three conservation principles for a dynamic system: en-
ergy, momentum, and moment of momentum. Every conservation principle is the
result of a symmetry in position and time. The conservation of energy indicates the
homogeneity of time, the conservation of momentum indicates the homogeneity in
position space, and the conservation of moment of momentum indicates the isotropy
in position space.
Proof Consider a mechanical system with fC degrees of freedom. Mathematically,
the dynamics of the system is expressed by a set of n = fC second-order differential
equations of n unknown generalized coordinates qi (t), i = 1, 2, . . . , n:
¨
qi = Fi (qi , qi , t) i = 1, 2, . . . , n
˙ (2.108)
The general solution of the equations contains 2n constants of integrals.
˙
qi = qi (c1 , c2 , . . . , cn , t)
˙ i = 1, 2, . . . , n (2.109)
qi = qi (c1 , c2 , . . . , c2n , t) i = 1, 2, . . . , n (2.110)

19. 2.2 Energy 69
To determine these constants and uniquely identify the motion of the system, it is
˙
necessary to know the initial conditions qi (t0 ), qi (t0 ), which specify the state of the
system at some given instant t0 :
˙
cj = cj q(t0 ), q(t0 ), t0 j = 1, 2, . . . , 2n (2.111)
˙ ˙
fj q(t), q(t), t = cj q(t0 ), q(t0 ), t0 (2.112)
Each of these functions fj is an integral of the motion and each ci is a constant of
the motion. An integral of motion may also be called a first integral, and a constant
of motion may also be called a constant of integral.
When an integral of motion is given,
˙
f1 (q, q, t) = c1 (2.113)
we can substitute one of the equations of (2.108) with the first-order equation of
q1 = f (c1 , qi , qi+1 , t) i = 1, 2, . . . , n
˙ ˙ (2.114)
and solve a set of n − 1 second-order and one first-order differential equations:
qi+1 = Fi+1 (qi , qi , t)
¨ ˙
i = 1, 2, . . . , n (2.115)
q1 = f (c1 , qi , qi+1 , t)
˙ ˙
If there exist 2n independent first integrals fj , j = 1, 2, . . . , 2n, then instead of solv-
ing n second-order equations of motion (2.108), we can solve a set of 2n algebraic
equations
˙ ˙
fj (q, q) = cj q(t0 ), q(t0 ), t0 j = 1, 2, . . . , 2n (2.116)
and determine the n generalized coordinates qi , i = 1, 2, . . . , n:
qi = qi (c1 , c2 , . . . , c2n , t) i = 1, 2, . . . , n (2.117)
Generally speaking, an integral of motion f is a function of generalized coordi-
˙
nates q and velocities q such that its value remains constant. The value of an integral
of motion is the constant of motion c, which can be calculated by substituting the
˙
given value of the variables q(t0 ), q(t0 ) at the associated time t0 .
Example 46 (A mass–spring–damper vibrator) Consider a mass m attached to a
spring with stiffness k and a damper with damping c. The equation of motion of the
system and its initial conditions are
mx + cx + kx = 0
¨ ˙ (2.118)
x(0) = x0 x(0) = x0
˙ ˙ (2.119)
Its solution is
x = c1 exp(s1 t) + c2 exp(s2 t) (2.120)

20. 70 2 Vibration Dynamics
Fig. 2.13 A planar pendulum
√ √
c− c2 − 4km c+ c2 − 4km
s1 = s2 = (2.121)
−2m −2m
˙
Taking the time derivative, we find x:
x = c1 s1 exp(s1 t) + c2 s2 exp(s2 t)
˙ (2.122)
˙
Using x and x, we determine the integrals of motion f1 and f2 :
x − xs2
˙
f1 = = c1 (2.123)
(s1 − s2 ) exp(s1 t)
x − xs1
˙
f2 = = c2 (2.124)
(s2 − s1 ) exp(s2 t)
Because the constants of integral remain constant during the motion, we can calcu-
late their value at any particular time such as t = 0:
x0 − x0 s2
˙ x0 − x0 s1
˙
c1 = c2 = (2.125)
(s1 − s2 ) (s2 − s1 )
Substituting s1 and s2 provides us with the constants of motion c1 and c2 :
√ √
c2 − 4km(cx0 + x0 c2 − 4km + 2mx0 ) ˙
c1 = 2 − 4km)
(2.126)
2(c
√ √
c2 − 4km(cx0 − x0 c2 − 4km + 2mx0 ) ˙
c2 = 2 − 4km)
(2.127)
2(c
Example 47 (Constraint and first integral of a pendulum) Figure 2.13(a) illus-
trates a planar pendulum. The free body diagram of Fig. 2.13(b) provides us with
two equations of motion:
x
mx = −T
¨ (2.128)
l
y
my = −mg + T
¨ (2.129)
l

21. 2.2 Energy 71
Eliminating the tension force T , we have one second-order equation of two vari-
ables:
yx + xy + gx = 0
¨ ¨ (2.130)
Because of the constant length of the connecting bar we have a constraint equation
between x and y:
x2 + y2 − l2 = 0 (2.131)
Having one constraint indicates that we can express the dynamic of the system by
only one generalized coordinate. Choosing θ as the generalized coordinate, we can
express x and y by θ , writing the equation of motion (2.130) as
g
¨
θ+ sin θ = 0 (2.132)
l
˙
Multiplying the equation by θ and integrating provides us with the integral of en-
ergy:
1 g
˙ ˙
f (θ, θ ) = θ 2 − cos θ = E (2.133)
2 l
1 2 g
˙
E = θ0 − cos θ0 (2.134)
2 l
The integral of motion (2.133) is a first-order differential equation:
g
˙
θ= 2E + 2 cos θ (2.135)
l
This equation expresses the dynamic of the pendulum upon solution.
Let us assume that θ is too small to approximate the equation of motion as
g
¨
θ + θ =0 (2.136)
l
The first integral of this equation is
1 g
˙ ˙
f (θ, θ ) = θ 2 − θ =E (2.137)
2 l
1 2 g
˙
E = θ0 − θ0 (2.138)
2 l
that provides us with a separated first-order differential equation:
g
˙
θ= 2E + 2 θ (2.139)
l
Its solution is
dθ √ l g
t= = 2 θ +E−p (2.140)
2E + 2gθ g l
l

22. 72 2 Vibration Dynamics
where p is the second constant of motion:
l
p= ˙
θ0 (2.141)
g
Now, let us ignore the energy integral and solve the second-order equation of
motion (2.136):
g g
θ = c1 cos t + c2 sin t (2.142)
l l
The time derivative of the solution
l g g
˙
θ = −c1 sin t + c2 cos t (2.143)
g l l
can be used to determine the integrals and constants of motion:
g l g
f1 = θ cos t− ˙
θ sin t (2.144)
l g l
g l g
f2 = θ sin t+ ˙
θ cos t (2.145)
l g l
˙ ˙
Using the initial conditions θ (0) = θ0 , θ (0) = θ0 , we have
l
c1 = θ 0 c2 = ˙
θ0 (2.146)
g
A second-order equation has only two constants of integrals. Therefore, we should
be able to express E and p in terms of c1 and c2 or vice versa:
1 2 g 1g 2 g
˙
E = θ0 − θ0 = c − c1 (2.147)
2 l 2l 2 l
l l g
p= ˙
θ0 = c2 (2.148)
g g l
l g l
c2 = ˙
θ0 = p (2.149)
g l g
1g 2 l
c1 = θ 0 = p − E (2.150)
2l g
E is the mechanical energy of the pendulum, and p is proportional to its moment of
momentum.

23. 2.3 Rigid Body Dynamics 73
Fig. 2.14 A globally fixed
G-frame and a body B-frame
with a fixed common origin
at O
2.3 Rigid Body Dynamics
A rigid body may have three translational and three rotational DOF. The transla-
tional and rotational equations of motion of the rigid body are determined by the
Newton–Euler equations.
2.3.1 Coordinate Frame Transformation
Consider a rotation of a body coordinate frame B(Oxyz) with respect to a global
frame G(OXY Z) about their common origin O as illustrated in Fig. 2.14. The com-
ponents of any vector r may be expressed in either frame. There is always a trans-
formation matrix G RB to map the components of r from the frame B(Oxyz) to the
other frame G(OXY Z):
G
r = G RB B r (2.151)
−1
In addition, the inverse map B r = G RB G r can be done by B RG ,
B
r = B RG G r (2.152)
where
G B
RB = RG = 1 (2.153)
and
−1
B
RG = G RB = G RB
T
(2.154)
When the coordinate frames B and G are orthogonal, the rotation matrix GR
B is
called an orthogonal matrix. The transpose R T and inverse R −1 of an orthogonal
matrix [R] are equal:
R T = R −1 (2.155)

24. 74 2 Vibration Dynamics
Because of the matrix orthogonality condition, only three of the nine elements of
G R are independent.
B
Proof Employing the orthogonality condition
ˆı ˆ ˆ ˆ ˆ
r = (r · ı )ˆ + (r · j )j + (r · k)k (2.156)
and decomposition of the unit vectors of G(OXY Z) along the axes of B(Oxyz),
ˆ ˆ ˆı ˆ ˆ ˆ ˆ ˆ ˆ
I = (I · ı )ˆ + (I · j )j + (I · k)k (2.157)
ˆ ˆ ˆı ˆ ˆ ˆ ˆ ˆ ˆ
J = (J · ı )ˆ + (J · j )j + (J · k)k (2.158)
ˆ ˆ ˆı ˆ ˆ ˆ ˆ ˆ ˆ
K = (K · ı )ˆ + (K · j )j + (K · k)k (2.159)
introduces the transformation matrix G RB to map the local axes to the global axes:
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤
Iˆ ˆ ˆ
I ·ı ˆ ˆ
I ·j ˆ ˆ
I ·k ˆ
ı ˆ
ı
⎣J ⎦ = ⎣J · ı
ˆ ˆ ˆ ˆ ˆ
J ·j J · k ⎦ ⎣j ⎦ = G RB ⎣j ⎦
ˆ ˆ ˆ ˆ (2.160)
ˆ
K ˆ ˆ
K ·ı ˆ ˆ
K ·j ˆ ˆ
K ·k ˆ
k ˆ
k
where
⎡ ⎤
ˆ ˆ ˆ ˆ
I ·ı I ·j I ·k ˆ ˆ
G
RB = ⎣ J · ı J · j J · k ⎦
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ·ı K ·j K ·k
K ˆ ˆ ˆ ˆ ˆ
⎡ ⎤
ˆ ˆ ˆ ˆ ˆ ˆ
cos(I , ı ) cos(I , j ) cos(I , k)
= ⎣ cos(J , ı ) cos(J , j ) cos(J , k) ⎦
ˆ ˆ ˆ ˆ ˆ ˆ (2.161)
ˆ ˆ ˆ ˆ ˆ ˆ
cos(K, ı ) cos(K, j ) cos(K, k)
Each column of G RB is the decomposition of a unit vector of the local frame
B(Oxyz) in the global frame G(OXY Z):
G
RB = ˆ
Gı ˆ
Gj ˆ
Gk (2.162)
Similarly, each row of G RB is decomposition of a unit vector of the global frame
G(OXY Z) in the local frame B(Oxyz).
⎡B ⎤
ˆ
IT
G
RB = ⎣ B J T ⎦
ˆ (2.163)
ˆ
B KT
so the elements of G RB are directional cosines of the axes of G(OXY Z) in
B(Oxyz) or B in G. This set of nine directional cosines completely specifies the
orientation of B(Oxyz) in G(OXY Z) and can be used to map the coordinates of
any point (x, y, z) to its corresponding coordinates (X, Y, Z).

25. 2.3 Rigid Body Dynamics 75
Alternatively, using the method of unit-vector decomposition to develop the ma-
trix B RG leads to
−1
r = B RG G r = G RB G r
B
(2.164)
⎡ ⎤
ˆ ˆ ˆ ˆ ˆ ˆ
ı ·I ı ·J ı ·K
B
RG = ⎣ j · I j · J j · K ⎦
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
k·I k·J k·K
⎡ ⎤
ı ˆ ı ˆ ı ˆ
cos(ˆ, I ) cos(ˆ, J ) cos(ˆ, K)
= ⎣cos(j , I ) cos(j , J ) cos(j , K)⎦
ˆ ˆ ˆ ˆ ˆ ˆ (2.165)
ˆ I ) cos(k, J ) cos(k, K)
cos(k, ˆ ˆ ˆ ˆ ˆ
It shows that the inverse of a transformation matrix is equal to the transpose of the
transformation matrix,
−1
G
RB = G RB
T
(2.166)
or
G
RB · G RB = I
T
(2.167)
A matrix with condition (2.166) is called an orthogonal matrix. Orthogonality of
GR
B comes from the fact that it maps an orthogonal coordinate frame to another
orthogonal coordinate frame.
An orthogonal transformation matrix G RB has only three independent elements.
The constraint equations among the elements of G RB will be found by applying the
matrix orthogonality condition (2.166):
⎡ ⎤⎡ ⎤ ⎡ ⎤
r11 r12 r13 r11 r21 r31 1 0 0
⎣r21 r22 r23 ⎦ ⎣r12 r22 r32 ⎦ = ⎣0 1 0⎦ (2.168)
r31 r32 r33 r13 r23 r33 0 0 1
Therefore, the inner product of any two different rows of G RB is zero, and the inner
product of any row of G RB by itself is unity:
r11 + r12 + r13 = 1
2 2 2
r21 + r22 + r23 = 1
2 2 2
r31 + r32 + r33 = 1
2 2 2
(2.169)
r11 r21 + r12 r22 + r13 r23 = 0
r11 r31 + r12 r32 + r13 r33 = 0
r21 r31 + r22 r32 + r23 r33 = 0

26. 76 2 Vibration Dynamics
These relations are also true for columns of G RB and evidently for rows and
columns of B RG . The orthogonality condition can be summarized by the equation
3
rij rik = δj k j, k = 1, 2, 3 (2.170)
i=1
where rij is the element of row i and column j of the transformation matrix G RB
and δj k is the Kronecker delta δij ,
1 i=j
δij = δj i = (2.171)
0 i=j
Equation (2.170) provides us with six independent relations that must be satisfied
by the nine directional cosines. Therefore, there are only three independent direc-
tional cosines. The independent elements of the matrix G RB cannot be in the same
row or column or any diagonal.
The determinant of a transformation matrix is equal to unity,
G
RB = 1 (2.172)
because of Eq. (2.167) and noting that
G G G G G G 2
RB · G RB =
T
RB · RB =
T
RB · RB = RB =1 (2.173)
ˆ ˆ ˆ
Using linear algebra and column vectors G ı , G j , and G k of G RB , we know that
G
RB = G ı ·
ˆ G ˆ
j × Gk
ˆ (2.174)
and because the coordinate system is right handed, we have ˆ
Gj ˆ
× G k = G ı and,
ˆ
therefore,
G
RB = G ı T · G ı = +1
ˆ ˆ (2.175)
Example 48 (Global position using B r and B RG ) The position vector r of a point
P may be described in either the G(OXY Z) or the B(Oxyz) frame. If B r = 10ˆ −ı
ˆ
5j + 15k and the transformation matrix to map G r to B r is
ˆ
⎡ ⎤
0.866 0 0.5
B
r = B RG G r = ⎣−0.353 0.707 0.612 ⎦ G r (2.176)
0.353 0.707 −0.612
then the components of G r in G(OXY Z) would be
⎡ ⎤
15.72
G
r = G RB B r = B RG B r = ⎣ 7.07 ⎦
T
(2.177)
−7.24

27. 2.3 Rigid Body Dynamics 77
Example 49 (Two-point transformation matrix) The global position vectors of two
points P1 and P2 , of a rigid body B are
⎡ ⎤ ⎡ ⎤
1.077 −0.473
G
rP1 = ⎣1.365⎦ G
rP2 = ⎣ 2.239 ⎦ (2.178)
2.666 −0.959
The origin of the body B(Oxyz) is fixed on the origin of G(OXY Z), and the points
P1 and P2 are lying on the local x- and y-axis, respectively.
ˆ ˆ
To find G RB , we use the local unit vectors G ı and G j ,
⎡ ⎤ ⎡ ⎤
Gr 0.338 Gr −0.191
= ⎣0.429⎦ = ⎣ 0.902 ⎦
P1 P2
G
ı= G
ˆ G
j= G
ˆ (2.179)
| rP1 | 0.838 | rP2 | −0.387
ˆ
to obtain G k:
⎡ ⎤
−0.922
Gˆ
k = ı × j = ⎣−0.029⎦
ˆ ˆ (2.180)
0.387
Hence, the transformation matrix G RB would be
⎡ ⎤
0.338 −0.191 −0.922
G
RB = G ı G j G k = ⎣0.429 0.902 −0.029⎦
ˆ ˆ ˆ (2.181)
0.838 −0.387 0.387
Example 50 (Length invariant of a position vector) Expressing a vector in different
frames utilizing rotation matrices does not affect the length and direction properties
of the vector. Therefore, the length of a vector is an invariant property:
G B
|r| = r = r (2.182)
The length invariant property can be shown as
G TG
|r|2 = G rTG r = RB B r RB B r = B rTG RB G RB B r
T
= B rTB r (2.183)
Example 51 (Multiple rotation about global axes) Consider a globally fixed point
P at
⎡ ⎤
1
G
r = ⎣2⎦ (2.184)
3
The body B will turn 45 deg about the X-axis and then 45 deg about the Y -axis.
An observer in B will see P at

28. 78 2 Vibration Dynamics
B
r = Ry,−45 Rx,−45 G r
⎡ ⎤ ⎡1 ⎤⎡ ⎤
cos −π 0 − sin −π 0 0 1
⎦ ⎢0 cos 4 sin −π ⎥ ⎣2⎦
4 4 −π
=⎣ 0 1 0 ⎣ 4 ⎦
sin −π 0 cos −π
4 4 0 − sin −π cos −π 3
4 4
⎡ ⎤⎡ ⎤ ⎡ ⎤
0.707 0.5 0.5 1 3.207
=⎣ 0 0.707 −0.707⎦ ⎣2⎦ = ⎣−0.707⎦ (2.185)
−0.707 0.5 0.5 3 1.793
To check this result, let us change the role of B and G. So, the body point at
⎡ ⎤
1
B
r = ⎣2⎦ (2.186)
3
undergoes an active rotation of 45 deg about the x-axis followed by 45 deg about
the y-axis. The global coordinates of the point would be
B
r = Ry,45 Rx,45 G r (2.187)
so
G
r = [Ry,45 Rx,45 ]TB r = Rx,45 Ry,45 B r
T T
(2.188)
Example 52 (Multiple rotations about body axes) Consider a globally fixed point P
at
⎡ ⎤
1
G
r = ⎣2⎦ (2.189)
3
The body B will turn 45 deg about the x-axis and then 45 deg about the y-axis.
An observer in B will see P at
B
r = RY,−45 RX,−45 G r
⎡ ⎤ ⎡1 ⎤⎡ ⎤
cos −π 0 sin −π 0 0 1
4 4 ⎢ −π −π ⎥
=⎣ 0 1 0 ⎦ ⎣0 cos 4 − sin 4 ⎦ ⎣2⎦
− sin −π 0 cos −π
4 4 0 sin −π cos −π 3
4 4
⎡ ⎤⎡ ⎤ ⎡ ⎤
0.707 0.5 −0.5 1 0.20711
=⎣ 0 0.707 0.707⎦ ⎣2⎦ = ⎣ 3.5356 ⎦ (2.190)
0.707 −0.5 0.5 3 1.2071
Example 53 (Successive rotations about global axes) After a series of sequential
rotations R1 , R2 , R3 , . . . , Rn about the global axes, the final global position of a
body point P can be found by
G
r = G RB B r (2.191)

29. 2.3 Rigid Body Dynamics 79
where
G
RB = Rn · · · R3 R2 R1 (2.192)
The vectors G r and B r indicate the position vectors of the point P in the global and
local coordinate frames, respectively. The matrix G RB , which transforms the local
coordinates to their corresponding global coordinates, is called the global rotation
matrix.
Because matrix multiplications do not commute, the sequence of performing ro-
tations is important and indicates the order of rotations.
Proof Consider a body frame B that undergoes two sequential rotations R1 and
R2 about the global axes. Assume that the body coordinate frame B is initially
coincident with the global coordinate frame G. The rigid body rotates about a global
axis, and the global rotation matrix R1 gives us the new global coordinate G r1 of
the body point:
G
r 1 = R1 B r (2.193)
Before the second rotation, the situation is similar to the one before the first rota-
tion. We put the B-frame aside and assume that a new body coordinate frame B1
is coincident with the global frame. Therefore, the new body coordinate would be
B1 r ≡ G r . The second global rotation matrix R provides us with the new global
1 2
position G r2 of the body points B1 r:
B1
r = R2 B1 r (2.194)
Substituting (2.193) into (2.194) shows that
G
r = R 2 R1 B r (2.195)
Following the same procedure we can determine the final global position of a body
point after a series of sequential rotations R1 , R2 , R3 , . . . , Rn as (2.192).
Example 54 (Successive rotations about local axes) Consider a rigid body B with
a local coordinate frame B(Oxyz) that does a series of sequential rotations R1 , R2 ,
R3 , . . . , Rn about the local axes. Having the final global position vector G r of a body
point P , we can determine its local position vector B r by
B
r = B RG G r (2.196)
where
B
R G = R n · · · R3 R 2 R 1 (2.197)
The matrix B RG is called the local rotation matrix and it maps the global coordi-
nates of body points to their local coordinates.

30. 80 2 Vibration Dynamics
Proof Assume that the body coordinate frame B was initially coincident with the
global coordinate frame G. The rigid body rotates about a local axis, and a local
rotation matrix R1 relates the global coordinates of a body point to the associated
local coordinates:
B
r = R1 G r (2.198)
If we introduce an intermediate space-fixed frame G1 coincident with the new posi-
tion of the body coordinate frame, then
G1
r ≡ Br (2.199)
and we may give the rigid body a second rotation about a local coordinate axis. Now
another proper local rotation matrix R2 relates the coordinates in the intermediate
fixed frame to the corresponding local coordinates:
B
r = R2 G1 r (2.200)
Hence, to relate the final coordinates of the point, we must first transform its global
coordinates to the intermediate fixed frame and then transform to the original body
frame. Substituting (2.198) in (2.200) shows that
B
r = R 2 R1 G r (2.201)
Following the same procedure we can determine the final global position of a body
point after a series of sequential rotations R1 , R2 , R3 , . . . , Rn as (2.197).
Rotation about the local coordinate axes is conceptually interesting. This is be-
cause in a sequence of rotations each rotation is about one of the axes of the local
coordinate frame, which has been moved to its new global position during the last
rotation.
2.3.2 Velocity Kinematics
Consider a rotating rigid body B(Oxyz) with a fixed point O in a reference frame
G(OXY Z), as shown in Fig. 2.15. We express the motion of the body by a time-
varying rotation transformation matrix between B and G to transform the instanta-
neous coordinates of body points to their coordinates in the global frame:
G
r(t) = G RB (t)B r (2.202)
The velocity of a body point in the global frame is
G
˙ ˙
v(t) = G r(t) = G RB (t)B r = G ωB G r(t) = G ωB × G r(t)
˜ (2.203)
where G ωB is the angular velocity vector of B with respect to G. It is equal to a
˙ ˆ
rotation with angular speed φ about an instantaneous axis of rotation u:

31. 2.3 Rigid Body Dynamics 81
Fig. 2.15 A rotating rigid
body B(Oxyz) with a fixed
point O in a global frame
G(OXY Z)
⎤⎡
ω1
ω = ⎣ω2 ⎦ = φ u
˙ˆ (2.204)
ω3
˜
The angular velocity vector is associated with a skew-symmetric matrix G ωB called
the angular velocity matrix,
⎡ ⎤
0 −ω3 ω2
ω = ⎣ ω3
˜ 0 −ω1 ⎦ (2.205)
−ω2 ω1 0
where
˜
G ωB ˙ ˙˜
= G R B G RB = φ u
T
(2.206)
The B-expression of the angular velocity is similarly defined:
T ˙
B
˜
G ωB = G RB G R B (2.207)
Employing the global and body expressions of the angular velocity of the body
˜ G˜
relative to the global coordinate frame, G ωB and B ωB , we determine the global and
body expressions of the velocity of a body point as
G
G vP = G ω B × G rP
G (2.208)
B
G vP = B ω B × B rP
G (2.209)
˜ G˜
The G-expression G ωB and B-expression B ωB of the angular velocity matrix
can be transformed to each other using the rotation matrix G RB :
˜
G ωB = G RB B ω B G RB
G˜
T
(2.210)
B
˜
G ωB = G RB G ω B G RB
T
G˜ (2.211)
They are also related to each other directly by
˜
G ωB
G
RB = G RB B ω B
G˜
T
(2.212)

32. 82 2 Vibration Dynamics
Fig. 2.16 A body fixed point
P at B r in the rotating body
frame B
G
RB G ω B = B ω B G RB
T
˜ G˜
T
(2.213)
The relative angular velocity vectors of relatively moving rigid bodies can be
done only if all the angular velocities are expressed in one coordinate frame:
n
0 ωn = 0 ω1 + 1 ω2 + 2 ω3 + · · · + n−1 ωn =
0 0 0 0
i−1 ωi (2.214)
i=1
˜ G˜
The inverses of the angular velocity matrices G ωB and B ωB are
˜ −1
G ωB ˙ −1
= G RB G R B (2.215)
B −1
˜
G ωB
˙ −1
= G R B G RB (2.216)
Proof Consider a rigid body with a fixed point O and an attached frame B(Oxyz)
as shown in Fig. 2.16. The body frame B is initially coincident with the global frame
G. Therefore, the position vector of a body point P at the initial time t = t0 is
G
r(t0 ) = B r (2.217)
and at any other time is found by the associated transformation matrix G RB (t):
G
r(t) = G RB (t)B r = G RB (t)G r(t0 ) (2.218)
The global time derivative of G r is
Gd Gd Gd
G
˙
v = Gr = G
r(t) = G
RB (t)B r = G
RB (t)G r(t0 )
dt dt dt
G ˙ G ˙
= RB (t) r(t0 ) = RB (t)B r
G
(2.219)
Eliminating B r between (2.218) and (2.219) determines the velocity of the global
point in the global frame:
G ˙
v = G RB (t)G RB (t)G r(t)
T
(2.220)

33. 2.3 Rigid Body Dynamics 83
˜
We denote the coefficient of G r(t) by G ωB
˜
G ωB ˙
= G R B G RB
T
(2.221)
and rewrite Eq. (2.220) as
G
v = G ωB G r(t)
˜ (2.222)
or equivalently as
G
v = G ωB × G r(t) (2.223)
where G ωB is the instantaneous angular velocity of the body B relative to the global
frame G as seen from the G-frame.
Transforming G v to the body frame provides us with the body expression of the
velocity vector:
T ˙
B
G vP = G RB G v = G RB G ω B G r = G RB G R B G RB G r
T T
˜ T
T ˙
= G RB G R B B r (2.224)
G˜
We denote the coefficient of B r by B ωB
T ˙
B
˜
G ωB = G RB G R B (2.225)
and rewrite Eq. (2.224) as
B
G vP = B ω B B rP
G˜ (2.226)
or equivalently as
B
G vP = B ω B × B rP
G (2.227)
where Bω is the instantaneous angular velocity of B relative to the global frame
G B
G as seen from the B-frame.
The time derivative of the orthogonality condition, G RB G RB = I, introduces an
T
important identity,
G ˙ T ˙T
R B G RB + G RB G R B = 0 (2.228)
˜ ˙
which can be used to show that the angular velocity matrix G ωB = [G RB G RB ] is
T
skew-symmetric:
G ˙T
RB G R B =
G ˙ T
R B G RB
T
(2.229)
Generally speaking, an angular velocity vector is the instantaneous rotation of
a coordinate frame A with respect to another frame B that can be expressed in or
seen from a third coordinate frame C. We indicate the first coordinate frame A by
a right subscript, the second frame B by a left subscript, and the third frame C by
a left superscript, C ωA . If the left super and subscripts are the same, we only show
B
the subscript.

34. 84 2 Vibration Dynamics
We can transform the G-expression of the global velocity of a body point P ,
Gv and the B-expression of the global velocity of the point P , B vP , to each other
P, G
using a rotation matrix:
B
G vP = B RG G vP = B RGG ωB G rP = B RGG ωB G RB B rP
˜ ˜
˙ ˙
= B RG G R B G RB G RB B r P = B RG G R B B r P
T
T ˙
= G RB G R B B r P = B ω B B r P = B ω B × B r P
G˜ G (2.230)
G
vP = G
RB B v P
G = G
G˜
RB B ω B B r P = G
G˜
T
RB B ω B G RB G r P
T ˙ ˙
= G RB G RB G R B G RB G r P = G R B G RB G r P
T T
B
= G ω B G r P = G ω B × G r P = G RB
˜ ω
G B
× B rP (2.231)
˜ G˜
From the definitions of G ωB and B ωB in (2.221) and (2.225) and comparing with
(2.230) and (2.231), we are able to transform the two angular velocity matrices by
˜
G ωB = G RB B ω B G RB
G˜
T
(2.232)
B
˜
G ωB = G RB G ω B G RB
T
˜ (2.233)
and derive the following useful equations:
G ˙
R B = G ω B G RB
˜ (2.234)
G ˙
RB = G
G˜
RB B ω B (2.235)
˜
G ωB
G
RB = G RB B ω B
G˜ (2.236)
The angular velocity of B in G is negative of the angular velocity of G in B if
both are expressed in the same coordinate frame:
G
˜
G ωB = −G ω G
B ˜
G
G ωB = −G ω G
B (2.237)
B
˜
G ωB = −B ω G
B˜
B
G ωB = −B ω G
B (2.238)
The vector G ωB can always be expressed in the natural form
G ωB = ωu
ˆ (2.239)
ˆ
with the magnitude ω and a unit vector u parallel to G ωB that indicates the instan-
taneous axis of rotation.
To show the addition of relative angular velocities in Eq. (2.214), we start from
a combination of rotations,
0
R2 = 0 R1 1 R2 (2.240)
and take the time derivative:
0 ˙ ˙ ˙
R 2 = 0 R 1 1 R2 + 0 R1 1 R 2 (2.241)